## Calculus with Applications (10th Edition)

Horizontal lines have slope equal to 0, so to determine the values of $x$ where the tangent line is horizontal, we set the derivative of $f$ equal to $0$ and solve for $x$. Using the Power Rule, the derivative of $$f(x)= 2x^3 +9x^2-60x+4$$ is $$f'(x) = 6x^2+18x-60.$$ We set the derivative $f'(x)$ equal to 0 and solve for $x$: $$f'(x) = 6x^2+18x-60=0.$$ Factor out a 6 from the left side: $$6( x^2+3x-10)=0.$$ Dividing both sides of the equation by 6 gives: $$x^2+3x-10=0.$$ We factor the quadratic equation to obtain: $$(x+5)(x-2)=0.$$ Setting each factor equal to $0$ and solving for $x$, we get that $x=-5$ or $x=2$.